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\title{Lancaster Molecular Electronic Transport code v0.0}
\author{David Zsolt Manrique}
\date{2009}
\begin{document}


\maketitle
\section{Intorduction}
Lancaster Molecular Electronic Transport code is a program that can help to calculate the electronic transport properties of a TB molecule.
\section{Usual electronic transport setup}
A very basic setup is an infinite long wire with periodic structure and in a localized range there is a scatterer object. It can be generalized in a way that at the two sides of the scatterer range there may be two different wire and also one can add more than two wires eventually.
The Hamiltonian for the basic setup has the structure:
\begin{equation}
\mathcal{H}=
\left(
  \begin{array}{ccccccc}
    \dots & \dots & \,         & \,         & \,         & \, & \, \\
    \dots & H_1   & V_1        & \,         & \,         & \, & \, \\
       \, & V_1^{\dagger} & H_1        & \Gamma_1   & \,         & \, & \, \\
       \, & \,    & \Gamma_1^{\dagger} & H_0        & \Gamma_2   & \, & \, \\
       \, & \,    & \,         & \Gamma_2^{\dagger} & H_2        & V_2 & \, \\
       \, & \,    & \,         & \,         & V_2^{\dagger}      & H_2 & \dots \\
       \, & \,    & \,         & \,         & \,      & \dots & \dots \\
  \end{array}
\right)
\end{equation}
where $H_1$ and $H_2$ represent the unit cell Hamiltonian of the left and right wire, respectively and $V_1$ and $V_2$ the couplings in the wire between the unit cells. Furthermore $H_0$ is the scatterer Hamiltonian and $\Gamma_1$ and $\Gamma_2$ the coupling between the scatterer and the wire. The dots at the top and bottom note that the wire structures repeat infinitely and the rest of the Hamiltonian is zero. Since this form is hard to generalize to more than two wires we rearrange in the following way.
\begin{equation}
\mathcal{H}=
\left(
  \begin{array}{ccccccc}
        H_0        & \Gamma_1   & \Gamma_2 &          &        & \\
        \Gamma_1^{\dagger} & H_1        &          & V_1      &        & \\
        \Gamma_2^{\dagger} &            & H_2      &          & V_2    & \\
                   & V_1^{\dagger}      &          & H_1      &        & \\
                   &            & V_2^{\dagger}    &          & H_2    & \dots \\
                   &            &          &          & \dots  & \dots \\
  \end{array}
\right)
\end{equation}
This form is already general, since one can add the third wire to introduce a $\Gamma_3$,$V_3$ and $H_3$ and insert these matrices by following the pattern. One must notice that basically the general form is
\begin{equation}
\mathcal{H}=
\left(
  \begin{array}{ccccccc}
        H_0      & \Gamma  &        &                \\
        \Gamma^{\dagger} & H       & V      &                \\
                 & V^{\dagger}      & H     & \dots          \\
                 &          & \dots & \dots          \\
  \end{array}
\right)
\end{equation}
which is also the trivial form considering the model that a finite region is connected to infinite periodic structure, and in this picture we do not care about how that infinite structure is constructed, like does it constructed from one piece of wire or built up by many periodic wires together? The fine structure is coded into the structure of $H$ $\Gamma$ and $V$, having them block diagonal form.
\section{Transport properties}
Let us consider the basic transport setup that we have an infinite wire with periodic structure and some scatterer. Before one would like to see what is the effect of the scatterer, it is necessary to know  what are the transport properties of the wire itself without the scatterer. The Hamiltonian of the wire takes the form of
\begin{equation}
\mathcal{H}=
\left(
  \begin{array}{ccccccc}
        \dots     & \dots  &        &   &               \\
        \dots & H       & V      &      &            \\
                 & V^{\dagger}      & H     &  V  &          \\
                 &          &  V^{\dagger}   &  H  & \dots   \\
                 &          &        & \dots & \dots          \\
  \end{array}
\right)
\end{equation}
To keep the generality we introduce the overlap matrix which include the cases when the basis set is not orthogonal.
\begin{equation}
\mathcal{S}=
\left(
  \begin{array}{ccccccc}
        \dots     & \dots  &        &   &               \\
        \dots & S       & T      &      &            \\
                 & T^{\dagger}      & H     &  T  &          \\
                 &          &  T^{\dagger}   &  S  & \dots   \\
                 &          &        & \dots & \dots          \\
  \end{array}
\right)
\end{equation}
Therefore the SE is
\begin{equation}
\left(
  \begin{array}{ccccccc}
        \dots     & \dots  &        &   &               \\
        \dots & H       & V      &      &            \\
                 & V^{\dagger}      & H     &  V  &          \\
                 &          &  V^{\dagger}   &  H  & \dots   \\
                 &          &        & \dots & \dots          \\
  \end{array}
\right)
\left(
  \begin{array}{ccccccc}
        \dots                 \\
        \ket{\psi_{n-1}}                  \\
        \ket{\psi_{n}}         \\
        \ket{\psi_{n+1}}           \\
        \dots              \\
  \end{array}
\right)
= E
\left(
  \begin{array}{ccccccc}
        \dots     & \dots  &        &   &               \\
        \dots & S       & T      &      &            \\
                 & T^{\dagger}      & H     &  T  &          \\
                 &          &  T^{\dagger}   &  S  & \dots   \\
                 &          &        & \dots & \dots          \\
  \end{array}
\right)
\left(
  \begin{array}{ccccccc}
        \dots                 \\
        \ket{\psi_{n-1}}                  \\
        \ket{\psi_{n}}         \\
        \ket{\psi_{n+1}}           \\
        \dots              \\
  \end{array}
\right)
\end{equation}
where $\ket{\psi_n}$ is the $n$-th block vector with the same dimension as $H$ has.
One can obtain its eigenstates by using discrete Bloch-theorem, that is $\ket{\psi_{n+1}}=e^{ika}\ket{\psi_n} $.
Because of the periodicity every equation is the same we can write in one line.
\begin{equation}
V^{\dagger} \ket{\psi_{n-1}} + H \ket{\psi_{n}} + V \ket{\psi_{n+1}} = E \left [ T^{\dagger} \ket{\psi_{n-1}} + S \ket{\psi_{n}} + T \ket{\psi_{n+1}} \right ]
\end{equation}
by Bloch-theorem
\begin{equation}
\left [ V^{\dagger} e^{-ika} + H + V e^{ika} \right ] \ket{\psi_{n}}  = E \left [ T^{\dagger} e^{-ika} + S + T e^{ika} \right ] \ket{\psi_{n}}
\end{equation}
and for all $k$ from the BZ gives a solution of the SE and it results the $E_i(k)$  band structure. The scattering theory question is if now one puts a scatterer into this wire and a Bloch-wave is coming in by energy $E$ how much is going to be reflected and transmitted on the scatterer. This is the way how the transport properties are defined. this implies that we have to know what is the Bloch-wave for a given energy. One must notice that solving the eigenvalue and obtain the band structure gives us the relation that is the inverse what we actually need. We need those Bloch-waves and $k$ wave-numbers that belong to the energy $E$. By reformulating the equation above we get a different problem
\begin{equation}
[ (V^{\dagger} - E T^{\dagger}) e^{-ika} + \underbrace{(H-E S)}_{H_E}  + \underbrace{(V - E T)}_{V_E} e^{ika}] \ket{k}  = 0
\end{equation}
i.e. what are those $k$ values and $\ket{k}$ vectors that satisfy the equation above.
The two separate problem in a more concise form
\begin{equation}
H_k \ket{E}  = E S_k \ket{E}
\end{equation}
\begin{equation}
[ V_E^{\dagger} e^{-ika} + H_E  + V_E e^{ika}] \ket{k}  = 0
\end{equation}
Let us write this in matrix form. First we introduce the following notation: $z=e^{-ika}$ general complex number and the corresponding $\ket{k}=\ket{z}$. Furthermore
\begin{equation}
H(z) = V_E^{\dagger} \frac{1}{z} + H_E  + V_E z.
\end{equation}
The task is to find such a $z$,$\ket{z}$ pair that satisfies $H(z)\ket{z}=0$. Certainly there are $2N$ pairs. Let $z_n$ and $\ket{z_n}$ such pair. Then
\begin{equation}
\ket{z_n} = \sum_{j=1}^{2N} w_j^{(n)} \ket{j}
\end{equation}
so in the equation
\begin{equation}
\sum_{j=1}^{2N} \bra{i} V_E^{\dagger} \ket{j}  \frac{1}{z_n} w_j^{(n)}  + \sum_{j=1}^{2N}\bra{i} H_E \ket{j}  w_j^{(n)}  + \sum_{j=1}^{2N}\bra{i} V_E\ket{j} z_n w_j^{(n)} = 0
\end{equation}
so
\begin{equation}
\sum_{j=1}^{2N} \sum_{m=1}^{2N}\bra{i} V_E^{\dagger} \ket{j}  \frac{1}{z_n} \delta_{nm} w_j^{(m)}  + \sum_{j=1}^{2N}\bra{i} H_E \ket{j}  w_j^{(n)}  + \sum_{j=1}^{2N}\sum_{m=1}^{2N} \bra{i} V_E\ket{j} z_n \delta_{nm} w_j^{(m)} = 0
\end{equation}
By introducing $Z$ matrix such that $\bra{m} Z \ket{n} = z_n \delta_{nm}$ and introducing $W$ matrix that $\bra{j} W \ket{m} = w_j^{(m)}$
\begin{equation}
\sum_{j=1}^{2N} \sum_{m=1}^{2N} \bra{i} V_E^{\dagger} \ket{j} \bra{j} W \ket{m} \bra{m} Z^{-1} \ket{n}   + \sum_{j=1}^{2N} \bra{i} H_E \ket{j} \bra{j} W \ket{n}  + \sum_{j=1}^{2N} \sum_{m=1}^{2N} \bra{i} V_E\ket{j} \bra{j} W \ket{m} \bra{m} Z \ket{n}  = 0
\end{equation}
therefore
\begin{equation}
V_E^{\dagger}  W  Z^{-1}   +  H_E  W   +  V_E W  Z  = 0
\end{equation}
The solution is separated to two parts, $W_+$,$Z_+$ and $W_-$,$Z_-$ $N$ by $N$ matrices. ...

The Green function $\mathcal{G}$ of the wire is defined by
\begin{equation}
(E\mathcal{S}-\mathcal{H})\mathcal{G}=\mathcal{I}
\end{equation}
Since the wire is translation invariant $\mathcal{G}$ will be translation invariant too, in other words it has the same periodicity that $\mathcal{H}$ has, therefore we can write it in block form. (here there is an error, $H_E$ and $V_E$ was defined with the wrong sign. From now $H_E=E S - H$. So for the $n$-th row and column in eq up:
\begin{equation}
V_E^{\dagger} \mathcal{G}_{n-1,n} + H_E \mathcal{G}_{n,n} + V_E \mathcal{G}_{n+1,n}=I
\end{equation}
The general Green function is searched in the form of
\begin{equation}
\mathcal{G}_{i,j} = \sum_{n=1}^N \ket{z_{n,\mathcal{P}_{i-j}}} z_{n,\mathcal{P}_{i-j}}^{i-j} \bra{x_{n,\mathcal{P}_{i-j}}}
\end{equation}
where $\mathcal{P}_{\pm i}=\pm$ for all $i$ positive integer. We can use only one index because $\mathcal{G}_{i,j}=\mathcal{G}_{i-j}$
\begin{equation}
\mathcal{G}_{\alpha} = \sum_{n=1}^N \ket{z_{n,\mathcal{P}_{\alpha}}} z_{n,\mathcal{P}_{\alpha}}^{\alpha} \bra{x_{n,\mathcal{P}_{\alpha}}}
\end{equation}
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Since (?????????!! for all k??? what if complex??? is that solution of the SE?)$H(z)=H(z)^{\dagger}$ it follows that
\begin{equation}
H(z) = V_E^{\dagger} \frac{1}{z} + H_E  + V_E z = V_E \frac{1}{z^*} + H_E  + V_E^+ z^* = H\left(\frac{1}{z^*}\right)
\end{equation}
and since $H(z)\ket{z}=0$ therefore $H\left(\frac{1}{z^*}\right)\ket{z}=0$ and it follows $\ket{z}=\ket{\frac{1}{z^*}}$. Generally the possible $z$ values can be obtained from $\det H(z) = 0$ equation that gives $N$-th order polynomial function in $z$ and in $\frac{1}{z}$ that generally leads to $2N$-th order equation which gives $2N$ complex roots. However we know that the roots are in $z$,$\frac{1}{z^*}$ pairs therefore significant information are in $N$ roots(WRONG!). Where $N$ is the rank of the matrix $V_E$. (proof?) Let us assume we found $N$ $z$,$\ket{z}$ pairs that satisfy the eq () and among them no two $z$ and $z'$ that $z^*z'=1$. If $|z| = 1$ then it gives a propagating solution for the whole system because $k$ is real, in any other cases there is always a decaying factor that makes the solution decaying.


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